Expert: Paul Klarreich - 3/3/2006

Question
Question: I wanted to see how to derive the formula for the volume of a circular torus.  I know the volume of a circular torus is V=2(Pi)(squared)*radius(squared)*radius of cross section.  Thus, I wanted to see how this equation was derived.



1. Ashwin Kumar
2. Studing how to derive different volume formulas.
3.I have tried to manipulate the surface area of the torus to try and find the volume, but it has not worked.  I also drew the torus on the x and y axis is hopes that it would illustrate the right way to go after the problem but that did not help either.  The last thing I tried was to find the integral of the volume of a circle, but this just confused me more.

Answer
Hi, Ashwin,
(Should I address you as Kumar?)


Your Question: I wanted to see how to derive the formula for the volume of a circular torus. I know the volume of a circular torus is V=2(Pi)(squared)*radius(squared)*radius of cross section. Thus, I wanted to see how this equation was derived.

1. Ashwin Kumar
2. Studing how to derive different volume formulas.
3.I have tried to manipulate the surface area of the torus to try and find the volume, but it has not worked. I also drew the torus on the x and y axis is hopes that it would illustrate the right way to go after the problem but that did not help either. The last thing I tried was to find the integral of the volume of a circle, but this just confused me more
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What you want to do is a 'volume of revolution' where you revolve some area, in this case a circle, about some line.  Assume the torus is formed in this way:

1. A circle is drawn around the origin, with radius r.
2. The circle is revolved about the line  x = - R.

[CHEATING METHOD FOR GETTING THE ANSWER:]
Now there is something, a variation on Cavelieri's principle, that says if an area A is moved through a distance d, the volume swept out is Ad.  So in this case you are taking a circle, A = pi r^2, and moving it through a distance which is  d = 2 pi R, so the volume is

V = pi r^2 (2 pi R), which gives your answer.

You might not like that; neither do I.  So here is a dull, pedestrian, workaholic method, found in any calculus book -- the method of cylindrical shells.  In this method, you take a small vertical element and rotate it, forming a thin cylindrical shell.

This shell has a volume equal to its circumference times its height times its thickness.  (The thickness is usually a dx or dy or something.)

So draw yourself this picture:

The circle with center at (0,0).
A thin vertical section at x, going up and down to the boundary of the circle.

The height of this section is 2y, and y = sqrt(r^2 - x^2).

When it is revolved, the radius of revolution is  R + x.  It's exactly R when x is zero, at the origin, smaller to the left, bigger to the right.

The circumference of the shell is  2 pi(R + x)
The thickness is dx, and we will integrate from x = -r to x = +r.

We're ready:

WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


     Circumf       height          Thck
    ----------- -----------------  ----
dV = 2 pi(R + x) 2 sqrt(r^2 - x^2)   dx

dV = 4 pi(R + x)  sqrt(r^2 - x^2) dx

dV = 4 pi [R sqrt(r^2 - x^2) + x sqrt(r^2 - x^2)] dx

Now there is a nice trick for the second integral.  It is an ODD function of x, meaning  f(-x) = - f(x).  Any time you integrate an odd function over a symmetric interval, like -r to r, it is zero.  So out it goes at small expense.

          {
V = 4 pi R | sqrt(r^2 - x^2) dx
          }

We do the sqrt(r^2 - x^2) by a trig-substitution:

x = r sin t
dx = r cos t
sqrt(...) = r cos t

And we are NOW integrating  t from zero to pi.

          {t=pi
V = 4 pi R | r cos t r cos t dt
          }t=0


             {t=pi
V = 4 pi Rr^2 | cos^2 t dt
             }t=0

We do cos^2 t by the old half-angle trick.


         1 + cos(2t)  
cos^2 t = ------------
              2

{  1 + cos(2t)
| ------------- dt =
}       2

 t    sin 2t  |t=pi
(--- + ------) |
 2      4     |t=0

= (pi/2 + sin(2pi)/4) - (0 + sin 0 /4)

= pi/2

Put that in and you have your  2 pi R r^2  formula.